Astone is dropped at t = 0. a second stone, with 5 times the mass of the first, is dropped from the same point at t = 180 ms. (a) how far below the release point is the center of mass of the two stones at t = 420 ms? (neither stone has yet reached the ground.) (b) how fast is the center of mass of the two-stone system moving at that time?

a. 0.379m

b.2.646 m/s

Step-by-step explanation:

Let mass of first stone=

Mass of second stone=

Both stones are in  free fall.

The first one free fall during

The second one free fall during

Both stone start motion  at the same point with initial velocity=0

a.We have to find the  distance between the release point  and  below the release point where is the center of mass of the two stones.

Substitute the values

Hence, center of mass  from the release point at distance=0.379 m

b.We have to find

Now,

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation:

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step-by-step explanation: